R函数“tapply”,“do.call”和“plyr”包的“laply”出错

时间:2018-04-13 16:24:12

标签: r apply sapply tapply do.call

我想将我的功能 “F.w” 应用于 “t” 的相应元素。我尝试使用 sapply tapply do.call laply ......每次我获得错误!我做错了什么?

R代码:

Data=data.frame(X1=c("A","B","B","B","B","B","B","B","B","B","B","B","A","B",
               "C","B","B","A","A","A","B","B","C","C","A","B","A","B"),
             X2=rep(0,28))
w=list(rep(0.2,7),rep(0.5,18),rep(0.8,3))
F.w <- function(j){
  i <- which(Data$X1== unique(Data$X1)[j])
  Data$X2[i] <- as.numeric(unlist(w[j]))
  return(Data)
}
t= c(1,2,3)

library(plyr)
laply(t,F.w)

do.call(F.w,list(t),quote = TRUE)

tapply(t,t,F.w)

使用do.call我收到了这个错误:

  

警告信息:1:在is.na(e1)|中is.na(e2):较长的物体长度   不是较短物体长度2的倍数:在==.default中(数据$ X1,   unique(数据$ X1)[j]):较长的对象长度不是倍数   较短的物体长度3:在Data $ X2 [i]&lt; - as.numeric(unlist(w [j]))中:
  要替换的项目数量不是替换长度的倍数

预期结果

>Data
   X1  X2
1   A 0.2
2   B 0.5
3   B 0.5
4   B 0.5
5   B 0.5
6   B 0.5
7   B 0.5
8   B 0.5
9   B 0.5
10  B 0.5
11  B 0.5
12  B 0.5
13  A 0.2
14  B 0.5
15  C 0.8
16  B 0.5
17  B 0.5
18  A 0.2
19  A 0.2
20  A 0.2
21  B 0.5
22  B 0.5
23  C 0.8
24  C 0.8
25  A 0.2
26  B 0.5
27  A 0.2
28  B 0.5

1 个答案:

答案 0 :(得分:1)

F.w功能可以更改为

F.w <- function(dat, w1){
  i <- match(dat[["X1"]], unique(dat[["X1"]]))
  unlist(w1)[ rank(i, ties.method = "first")]
}

Data$X2 <-  F.w(Data, w)
Data$X2
#[1] 0.2 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.2 0.5 0.8 0.5 0.5
#[18] 0.2 0.2 0.2 0.5 0.5 0.8 0.8 0.2 0.5 0.2 0.5